Let `f(x)=int x^2/((1+x^2)(1+sqrt(1+x^2)))dx` and `f(0)=0` then `f(1)` is

Options

A. `log(1+sqrt(2))`
B. `log(1+sqrt(2))-(pi)/(4)`
C. `log(1+sqrt(2))+(pi)/(4)`
D. none of these

Answer

B

Explanation

`f(x)=int(x^(2)dx)/((1+x^(2))(1+sqrt(1+x^(2))))`
let `x=tan theta`
`implies dx=sec^(2) theta d theta=(1+x^(2)).d theta`
`impliesf(x)=int(x^(2)dx)/((1+x^(2))(1+sqrt(1+x^(2))))=int(tan^(2) theta sec^(2) theta d theta)/(sec^(2) theta(1+sec theta))`
`=int(tan^(2)theta d theta)/(1+sec theta)`
`=int (sin^(2)theta d theta)/(cos theta(1+cos theta))`
`=int(1-cos^(2)theta d theta)/(cos theta(1+cos theta))`
`=int ((1-cos theta)d theta)/(cos theta)`
`=int sec theta d theta-int d theta`
`=log(x+sqrt(1+x^(2)))-tan^(-1)x+c`
Given `f(0)=0`
`implies 0=log 1-0+c`
`implies c=0`
`implies f(1)=log(1+sqrt(1+1^(2)))-tan^(-1)(1)`
`=log(1+sqrt(2))-(pi)/(4).`
`f'(x)=(sqrt(1+x^(2))-1)/((1+x^(2))) gt 0` for all real x.
So, `f(x)` is an increasing function.